Simulating Halt Decider Theorem (Olcott 2021)

[olcott]

https://en.wikipedia.org/wiki/Halting_problem

This halt deciding criteria refutes all of the conventional halting
problem proofs because a simulating halt decider would decide that all
of the conventional halting problem counter-examples are infinitely
recursive.

Simulating Halt Decider Theorem (Olcott 2021):
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.

Halting problem undecidability and infinitely nested simulation
https://www.researchgate.net/publication/351947980_Halting_problem_undecidability_and_infinitely_nested_simulation


--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

[olcott]

On 10/25/2021 5:55 PM, dklei...@gmail.com wrote:
> On Monday, October 25, 2021 at 1:10:55 PM UTC-7, olcott wrote:
>> https://en.wikipedia.org/wiki/Halting_problem

>>
>> Simulating Halt Decider Theorem (Olcott 2021):
>> Whenever simulating halt decider H correctly determines that input P
>> never reaches its final state (whether or not its simulation of P is
>> aborted) then H correctly decides that P never halts.
>

> All this says is that if P never reaches a final state P never halts.
> That is just the definition of "halts".

q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn

https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf

Sure that seems to be totally obvious and no big deal until one realizes
that when the Linz Ĥ is applied to ⟨Ĥ⟩ (its own TM description) that the
simulating halt decider at Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly determines that its
input never halts. All of the conventional proofs (and Linz himself)
conclude that this is impossible.

>
> PS: Nit to be fixed: there can be more than one final attae.

[olcott]

On 10/25/2021 6:31 PM, André G. Isaak wrote:

> On 2021-10-25 17:11, olcott wrote:
>> On 10/25/2021 5:55 PM, dklei...@gmail.com wrote:
>>> On Monday, October 25, 2021 at 1:10:55 PM UTC-7, olcott wrote:
>>>> https://en.wikipedia.org/wiki/Halting_problem
>>>>
>>>> Simulating Halt Decider Theorem (Olcott 2021):
>>>> Whenever simulating halt decider H correctly determines that input P
>>>> never reaches its final state (whether or not its simulation of P is
>>>> aborted) then H correctly decides that P never halts.
>>> All this says is that if P never reaches a final state P never halts.
>>> That is just the definition of "halts".
>>
>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>>
>> https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>
>> Sure that seems to be totally obvious and no big deal until one
>> realizes that when the Linz Ĥ is applied to ⟨Ĥ⟩ (its own TM
>> description) that the simulating halt decider at Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
>> determines that its input never halts. All of the conventional proofs
>> (and Linz himself) conclude that this is impossible.
>

> Except that when we actually *run* Ĥ ⟨Ĥ⟩ it halts. Yes your decider
> claims it doesn't halt, but your decider is *wrong*.
>
> André
>

It can't possibly be wrong because it is a semantic tautology.
If we know that X is a black then then we know that X is a cat.

THIS STATEMENT IS NECESSARILY ALWAYS TRUE

Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.

[Jeff Barnett]

On 10/25/2021 5:57 PM, olcott wrote:

> It can't possibly be wrong because it is a semantic tautology.
> If we know that X is a black then then we know that X is a cat.

Your illness is turning your brain to jello. Read the above. No, No!
Read it again! It makes about as much sense as what else you drop on
this list. See your doctor, take your meds. Pursuing intellectual
endeavors is not for you. If you want to be remembered fondly, do a nice
thing or two for some disadvantaged child. Just don't try to impress
everyone with your thinking. If it were anyone else, I'd think it was a
typo but you have doubled down on nuttier crap than this on many occasions.
--
Jeff Barnett

[André G. Isaak]

On 2021-10-25 17:57, olcott wrote:
> On 10/25/2021 6:31 PM, André G. Isaak wrote:
>> On 2021-10-25 17:11, olcott wrote:
>>> On 10/25/2021 5:55 PM, dklei...@gmail.com wrote:
>>>> On Monday, October 25, 2021 at 1:10:55 PM UTC-7, olcott wrote:
>>>>> https://en.wikipedia.org/wiki/Halting_problem
>>>>>
>>>>> Simulating Halt Decider Theorem (Olcott 2021):
>>>>> Whenever simulating halt decider H correctly determines that input P
>>>>> never reaches its final state (whether or not its simulation of P is
>>>>> aborted) then H correctly decides that P never halts.
>>>> All this says is that if P never reaches a final state P never halts.
>>>> That is just the definition of "halts".
>>>
>>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>>>
>>> https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>>
>>> Sure that seems to be totally obvious and no big deal until one
>>> realizes that when the Linz Ĥ is applied to ⟨Ĥ⟩ (its own TM
>>> description) that the simulating halt decider at Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩
>>> correctly determines that its input never halts. All of the
>>> conventional proofs (and Linz himself) conclude that this is impossible.
>>
>> Except that when we actually *run* Ĥ ⟨Ĥ⟩ it halts. Yes your decider
>> claims it doesn't halt, but your decider is *wrong*.
>>
>> André
>>
>
> It can't possibly be wrong because it is a semantic tautology.
> If we know that X is a black then then we know that X is a cat.

We do? My TV remote is black. I guess it must be a cat...

>
> THIS STATEMENT IS NECESSARILY ALWAYS TRUE
> Whenever simulating halt decider H correctly determines that input P
> never reaches its final state (whether or not its simulation of P is
> aborted) then H correctly decides that P never halts.

But that is *not* the problem which a halt decider addresses.

q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn qn if M applied to <M> does not halt.

That last bit is what specifies the problem. Linz doesn't tell you how
to implement his H or Ĥ, but that exact same specification holds for a
simulating decider as much as it does for a non-simulating one.

You seem to be under the mistaken impression that how you implement your
H somehow allows you to change the specification. it doesn't.

As an example, consider a function which is specified to take an integer
x and return the an integer which is x doubled.

I can implement that in many different ways. I can return x + x. I can
return x * 2. I can return the result of a logical shift left. But in
all cases the specification remains "return the an integer which is x
doubled". I don't get to change it to 'return the result of a logical
shift left' just because that's how I've chosen to implement it.

Similarly, H/Ĥ must halt in a rejecting state if M applied to <M> does
not halt regardless of whether your H/Ĥ are simulating deciders or not.

And Ĥ applied to <Ĥ> HALTS.

André

--
To email remove 'invalid' & replace 'gm' with well known Google mail
service.

[olcott]

Sure it is. If H correctly decides that its input never halts then
nothing else in the universe can possibly show that H has decided its
input incorrectly.

[olcott]

On 10/25/2021 7:27 PM, André G. Isaak wrote:

> Inputs don't halt or not halt.
>
> And you *very* dishonestly cut the portion where I indicated the actual
> specification of the problem as defined by Linz.
>
> André
>

THIS IS TRUE BY LOGICAL NECESSITY:
If H correctly decides that its input represents a computation that

never halts then nothing else in the universe can possibly show that H

has decided the halt status of its input incorrectly.

[olcott]

On 10/25/2021 8:15 PM, Ben Bacarisse wrote:

> olcott <No...@NoWhere.com> writes:
>
>> On 10/25/2021 5:55 PM, dklei...@gmail.com wrote:
>>> On Monday, October 25, 2021 at 1:10:55 PM UTC-7, olcott wrote:
>>>> https://en.wikipedia.org/wiki/Halting_problem
>>>>
>>>> Simulating Halt Decider Theorem (Olcott 2021):
>>>> Whenever simulating halt decider H correctly determines that input P
>>>> never reaches its final state (whether or not its simulation of P is
>>>> aborted) then H correctly decides that P never halts.
>>> All this says is that if P never reaches a final state P never halts.
>>> That is just the definition of "halts".
>>
>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>>
>> https://www.liarparadox.org/Peter_Linz_HP(Pages_315-320).pdf
>>
>> Sure that seems to be totally obvious and no big deal
>

> Cut off from the specification is seems to be no big deal. Put the
> proper condition back:

>
> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn

> if (and only of) Ĥ applied to ⟨Ĥ⟩ does not halt
>
> and you can see why it is a big deal.

>
>> until one realizes that when the Linz Ĥ is applied to ⟨Ĥ⟩ (its own TM
>> description) that the simulating halt decider at Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly
>> determines that its input never halts.
>

> Declaring what Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ does "correct" does not make it so. What is
> correct follows from how Linz defines Ĥ, and is the part you keep lying
> by omitting.

>
>> All of the conventional proofs
>> (and Linz himself) conclude that this is impossible.
>

> All the proofs (conventional or otherwise) conclude that no TM can
> decide TM halting. Simply declaring that what Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ does is
> "correct" does not make it so.

>

THIS STATEMENT IS NECESSARILY ALWAYS TRUE

Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.

If the input to Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ never reaches is final state then Ĥq0 is
necessarily correct when its transitions to Ĥqn.

[olcott]

On 10/26/2021 9:23 AM, André G. Isaak wrote:
> On 2021-10-26 07:29, olcott wrote:
>> On 10/26/2021 12:55 AM, André G. Isaak wrote:
>>> On 2021-10-25 23:03, olcott wrote:
>>>> On 10/25/2021 11:38 PM, André G. Isaak wrote:
>>>>> On 2021-10-25 20:58, olcott wrote:
>>>
>>>>>> The problem with that way of saying it is that it is vague about
>>>>>> what point of the execution trace matters.
>>>>>
>>>>> There's nothing remotely vague about this. It states *exactly*
>>>>> under which circumstances Ĥq0 ⊢* Ĥqn is justified.
>>>>>
>>>>> The halting problem isn't about execution traces. Its about whether
>>>>> a particular TM halts when given a particular input string.
>>>>>
>>>>>> The precise way of saying that when we are assuming that Ĥq0 is a
>>>>>> simulating halt decider is:
>>>>>
>>>>> Assuming that Ĥq0 is a simulating halt decider makes not an iota of
>>>>> difference to the criterion which Linz states. Ĥq0 is supposed to
>>>>> transition to Ĥqn if and only if M applied to <M> does not halt.
>>>>
>>>> Which in my encoding means that the ⟨Ĥ⟩ applied to ⟨Ĥ⟩ simulated by Ĥq0
>>>
>>> No. It doesn't. It means exactly what it says. The specification of a
>>> problem doesn't change depending on the implementation.
>>>
>>> You can use whatever method you want internally, but if the final
>>> answer you get doesn't correspond to the one specified by the problem
>>> then your method simply isn't sound or you're answering the wrong
>>> question.
>>>
>>
>> Even if everyone else in the universe disagrees, it is always that
>> case that when a halt decider does correctly decide the halt status of
>> its input then this input did have its halt status correctly decided.
>> It seems a little psychotic that you disagree with this.
>
> People *aren't* disagreeing with this conditional (though it is
> horrendously worded).
>
> People are disagreeing with your assertion that your decider actually
> does decide correctly because it doesn't.

Simulating Halt Decider Theorem (Olcott 2021):
Whenever simulating halt decider H correctly determines that input P
never reaches its final state (whether or not its simulation of P is
aborted) then H correctly decides that P never halts.

q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn

As soon as I show that Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ meets the above theorem then Linz has
been refuted. Only a fool would disagree.

> To decide correctly you must
> actually match the condition specified by the problem, which is "does Ĥ
> applied to <Ĥ> halt". It does.
>
> André

[olcott]

On 10/26/2021 12:22 PM, André G. Isaak wrote:

> Calling something a theorem doesn't actually make it one. You should
> learn what 'theorem' means.
>

It is actually a self-evident truth, yet math, logic and computer
science people may have never heard of that term from philosophy so I
use its closest approximation: theorem

a general proposition not self-evident but proved by a chain of
reasoning; a truth established by means of accepted truths.

In epistemology (theory of knowledge), a self-evident proposition is a
proposition that is known to be true by understanding its meaning
without proof. https://en.wikipedia.org/wiki/Self-evidence

>> Whenever simulating halt decider H correctly determines that input P
>> never reaches its final state (whether or not its simulation of P is
>> aborted) then H correctly decides that P never halts.
>>
>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>
>> As soon as I show that Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ meets the above theorem then Linz
>> has been refuted. Only a fool would disagree.
>

> Things don't 'meet a theorem'. That's gibberish. They satisfy a
> criterion. And in the case of Linz's H/Ĥ the criterion is ALREADY STATED
> as part of the problem (though you keep deleting it).
>
> Linz CLEARLY states the criterion:
>
> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn IFF Ĥ APPLIED TO ⟨Ĥ⟩ HALTS.
>

If Linz says this then Linz is wrong. A halt decider is correct iff it
correctly decides the halt status of it inputs.

The halt decider is at Ĥq0 it is not at q0.
The inputs to the halt decider at Ĥq0 are: ⟨Ĥ⟩ ⟨Ĥ⟩.

As long as Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly determines that its inputs never reach
their final state then Ĥq0 is necessarily correct and impossibly incorrect.

> You can't replace the last bit with some other criterion and still claim
> to be working on the same problem.

[olcott]

On 10/26/2021 2:11 PM, André G. Isaak wrote:

> Anyone who has studied math or logic will have a far better grasp of
> philosophy than you do. I have no idea where you get the idea that
> people in these fields are ignorant of the relevant philsophy.
>
> And 'self-evident truth' isn't a notion that has played a significant
> role in philosophy in at least the last century and a half. It's an
> antiquated notion and it doesn't even remotely approximate 'theorem'.
>

There you go, your only understanding of philosophy is learned-by-rote.
Because you and all other logicians and mathematicians really only have
learned-by-rote as your basis you have no deep philosophical
understanding that self-evident truth provides the ultimate foundation
of all analytical knowledge.

> But the main point is that whatever you want to call this it isn't
> actually the criterion which is specified by the problem at hand. If you
> want to address the halting problem, you need to use the same criterion
> as that problem.

>
>> a general proposition not self-evident but proved by a chain of
>> reasoning; a truth established by means of accepted truths.
>>
>> In epistemology (theory of knowledge), a self-evident proposition is a
>> proposition that is known to be true by understanding its meaning
>> without proof. https://en.wikipedia.org/wiki/Self-evidence
>

> A notion dating back to the Ancient Greeks which has proven to be
> entirely inadequate. Euclid considered his fifth postulate to be
> self-evident[*]. But no one today would accept it as such.

>
>>
>>>> Whenever simulating halt decider H correctly determines that input P
>>>> never reaches its final state (whether or not its simulation of P is
>>>> aborted) then H correctly decides that P never halts.
>>>>
>>>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn
>>>
>>>> As soon as I show that Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ meets the above theorem then Linz
>>>> has been refuted. Only a fool would disagree.
>>>
>>> Things don't 'meet a theorem'. That's gibberish. They satisfy a
>>> criterion. And in the case of Linz's H/Ĥ the criterion is ALREADY
>>> STATED as part of the problem (though you keep deleting it).
>>>
>>> Linz CLEARLY states the criterion:
>>>
>>> q0 ⟨Ĥ⟩ ⊢* Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥqn IFF Ĥ APPLIED TO ⟨Ĥ⟩ HALTS.
>>>
>>
>> If Linz says this then Linz is wrong. A halt decider is correct iff it
>> correctly decides the halt status of it inputs.
>

> A halt decider decides the halting status of the *computation*
> represented by its input. Which is exactly what Linz states. The input
> itself is just a string. It has no halting status.

>
>> The halt decider is at Ĥq0 it is not at q0.
>> The inputs to the halt decider at Ĥq0 are: ⟨Ĥ⟩ ⟨Ĥ⟩.
>

> Neither Ĥ nor Ĥq0 are halt deciders at all.
>

Linz says that Ĥq0 is the "no" path of H.
If you want to get freaking nit picky it is a non-halting determiner.

>> As long as Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly determines that its inputs never
>> reach their final state then Ĥq0 is necessarily correct and impossibly
>> incorrect.
>

> The inputs don't have final states.

>> As long as Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ correctly determines that

THE SIMULATION OF

>> its inputs never
>> reach their final state then Ĥq0 is necessarily correct and
>> impossibly
>> incorrect.

> The machine they represent does. And
> its about that machine applied to the supplied input string that a halt
> decider is expect to decide. That's a simple matter of definition.
>

If for whatever reason the behavior of Ĥ applied to ⟨Ĥ⟩ does not match
the correct halt status of the input to Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ it cannot possibly
show that this correct halt status is incorrect.

As long as Ĥq0 ⟨Ĥ⟩ ⟨Ĥ⟩ applies correct axioms to the behavior of the
simulation of its input such that it correctly determines that this
simulation of this input would never reach its final state whether or
not Ĥq0 aborts this simulation, then Ĥ is necessarily correct when it
transitions to Ĥn no matter what the behavior of Ĥ applied to ⟨Ĥ⟩ is.

> André
>
> [*] Or at least he presented it as such, though there are reasons to
> believe he wasn't overly satisfied with this and would rather have
> derived this from the first four postulates, but this proved impossible.
> Of course, he's dead, so we can't really ask him about this.